Question

A cell of e.m.f. $\epsilon$ and internal resistance $r$ sends current $1.0A$ when it is connected to an external resistance $1.9\Omega$. But it sends current $0.5A$ when it is connected to an external resistance $3.9\Omega$. Calculate the values of $\epsilon$ and $r$.

• A. $\epsilon =2.0V,r=0.1\Omega$
• B. $\epsilon =.1V,r=2.0\Omega$
• C. $\epsilon =.2V,r=1.0\Omega$
• D. Data insufficient

Answer 1

The correct option is A $\epsilon =2.0V,r=0.1\Omega$

When the circuit is closed, the resulting current not only flows through the external circuit, but through the source (cell) itself. The cell have an internal resistance, which causes an internal voltage drop, slightly reducing the voltage across the terminals. The larger the current, the larger the internal voltage drop, and the lower the terminal voltage.
The current in the circuit is calculated from the formula $I=\frac{\epsilon }{R+r}$ where,
$\epsilon$- emf of the cell
R- external resistance
r- internal resistance of the cell.
When the cell is connected to the resistance of 1.9 ohms the current relation is given as
That is, $1.0=\frac{e}{1.9+r}.Thatis,aftersimplification,e-r=1.9-eqn1$.
When the cell is connected to the resistance of 3.9 ohms the current relation is given as
That is, $0.5=\frac{e}{3.9+r}.Thatis,aftersimplification,e-0.5r=1.95-eqn2$.
Solving the linear eqn 1 and eqn 2 for the variables e and r, we get e=2.0 V.
Substituting the value of e in eqn 1, the variable r is determined to be 0.1 ohms.
Hence, the emf and the internal resistance of the cell are 2.0 V and 0.1 ohms respectively.